Consider an open set on the plane $O\subseteq \mathbb{R}^2$ which is both bounded and simply connected, and such that $O$ is exactly the interior of its closure. Is the boundary $\partial O$ necessarily path connected?
I know that if we remove the boundedness condition, this fails when taking $O$ as infinite open strip, which has disconnected boundary. If we remove the simply simply connected condition, we can instead use an annulus to similar effect. If we remove the regular condition, we can use an open ball with the closed topologist's sine curve cut out. Are all of these conditions combined enough though?
Best Answer
No. Consider this modified version of the Warsaw circle, which has two instances of the Topologist's sine curve on it instead of just one. This shape can be given in polar coordinates as $r(\theta)=\cos\left(e^{\tan(\theta)}\right)+2$. Taking the open set $O$ as the interior region cut out by this curve, $O$ is bounded, simply connected, and regular, but its boundary is the closed image of this modified warsaw circle, which is not path connected. There is no way to path connect across the singularities at $\theta=\pm\frac{\pi}{2}$, so the boundary is path disconnected.